Yeah... I know it's sad, but I don't understand it, I'm doing math by correspondence and I don't have anyone to help me with it, I've watched up like 10 videos and at least read four guides but I still don't get it.
I've been in this calculus for 2 weeks now and I figured I skip it and coast off everything else, however, I got an assignment
for example I don't even know how to start with this question:
Use mathematical induction to prove that each formula is valid for all positive integral values of n.
-(1/2)-(1/4)-(1/8)-....-(1/2n)=1/2n-1 n∈N
I can do everything until the (k+1)=n part, I understand that K+1 is supposed represent the next part in the sequence algebraically but I get confused on how to show that and simplify.
Let's take this in two parts.
First, the concept. There is something we want to prove, P, about all integers \(\displaystyle \ge\) j. So we first prove that P is true for j. Now, because it is true for j, that shows that there are one or more integers that are equal or larger than j for which P is true. With me so far? Now we choose an arbitrary one of those integers, k. But that means that P is true for k. This is all we know about k: it is an integer, P is true about it, and it is not less than j. We can use any or all those three facts in our proof. We now prove that P is true of k + 1, which is also an integer not less than j. We then say axiomatically P is true for all integers not less than j. Why does the axiom make sense? Let's say that j = 1. By our first step in the proof, we proved P true for 1. By our second step in the proof, we have shown that P is true for 1 + 1 = 2, but that means it is true for 2 + 1 = 3, which in turn makes it true for 3 + 1 = 4, and consequently true for 4 + 1 = 5, and so on without end. Is this concept clear in your mind?
Second, mechanics. With respect to proving the first step, this is usually not difficult, but it may require creativity. The second step is usually much tougher. There may be many ways to proceed, but one that I have frequently found helpful is this.
Let's say P is of the the form f(n) = g(n). Now P is true of k so f(k) = g(k).
Express f(k + 1) in terms of k. Express g(k + 1) in terms of k. They should be equal in terms of k. This is very abstract so I have an example. Look out for typos.
Prove \(\displaystyle If\ n\ \in \mathbb Z\ and\ n \ge 1,\ then\ \displaystyle \sum_{i=1}^n(2i-1)^2 = \dfrac{4n^3 - n}{3}.\) j is 1.
Step 1: \(\displaystyle \displaystyle \sum_{i=1}^1(2i-1)^2 = 1^2 = 1 = \dfrac{3}{3} = \dfrac{4 - 1}{3} = \dfrac{4 * 1^3 - 1}{3}.\) True of 1.
Step 2: \(\displaystyle \exists\ k\ such\ that\ k \in \mathbb Z,\ k \ge 1,\ and\ \displaystyle \sum_{i=1}^k(2i-1)^2 = \dfrac{4k^3 - k}{3}.\) With me to here?
Now I want to express \(\displaystyle \displaystyle \sum_{i=1}^{k+1}(2i-1)^2\ in\ terms\ of\ k.\)
That is not too hard because:
\(\displaystyle \displaystyle \sum_{i=1}^{k+1}(2i-1)^2 = \{2(k + 1) - 1\}^2 + \sum_{i=1}^k(2i-1)^2 = (2k + 1)^2 +\dfrac{4k^3 - k}{3} = 4k^2 + 4k + 1 + \dfrac{4k^3 - k}{3} = \dfrac{4k^3 + 12k^2 + 11k + 3}{3}.\)
Next I want to express \(\displaystyle \dfrac{2(k + 1)^3 - (k + 1)}{3}\ in\ terms\ of\ k.\)
This too is just algebra:
\(\displaystyle \dfrac{4(k + 1)^3 - (k + 1)}{3} = \dfrac{4(k^3 + 3k^2 + 3k + 1) - k - 1}{3} = \dfrac{4k^3 + 12k^2 + 11k + 3}{3}.\)
\(\displaystyle \displaystyle \sum_{i=1}^{k+1}(2i-1)^2 = \dfrac{4k^3 + 12k^2 + 11k + 3}{3} = \dfrac{4(k + 1)^3 - (k + 1)}{3} \implies\)
\(\displaystyle \displaystyle \sum_{i=1}^{k+1}(2i-1)^2 = \dfrac{4(k + 1)^3 - (k + 1)}{3} \implies\)
\(\displaystyle n \in \mathbb \ Z\ and\ n \ge 1 \implies \displaystyle \sum_{i=1}^n(2i-1)^2 = \dfrac{4n^3 - n}{3}.\ QED.\)
I do not guarantee that this is the most efficient method, but it works often enough for me.
So give your problem a try, and show us your work if you get stuck. It comes in time. You'll get it.
